GATE CSE 2000 | Question: 4

Question

Let $S= \{0, 1, 2, 3, 4, 5, 6, 7\}$ and $⊗$ denote multiplication modulo $8,$ that is, $x ⊗ y= (xy) \mod 8$

1. Prove that $( \{ 0, 1\}, ⊗)$ is not a group.
2. Write three distinct groups $(G, ⊗)$ where $G ⊂ S$ and $G$ has $2$ elements.

Comments

a) 0 inverse does not exists for the given group so it is not a group

b) To make proper subgroup of 2 elements:

we know that 1 belongs to all the subgroups of the given group. Now we have to choose one more element.

So now choose that element which is coprime to 8 because that element is known as unit and inverse exist for that element. those elements are 3,5,7

So three distinct proper subgroups with cardinality 2 are:

{1,3} ,{1,5}, {1,7}

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@Vivek jangir such a simple yet nice explaination

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S itself is not a group because 0,2,4,6 don't have inverse

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thats ok , we can select some element from that and make group ,

U8 = {1,3,5,7} is a group with no generators , so it is not cyclic .

follow go classes for more detail.

Answer 1

$\begin{array}{c|cc}
⊗&0&1\\ \hline
0&0&0\\
1&0&1
\end{array}$

a. $1$ is the identity element. Inverse does not exist for zero. So, it is not a group.

b. $\begin{array}{c|cc}
⊗&1&3\\ \hline
1&1&3\\
3&3&1
\end{array}\qquad\begin{array}{c|cc}
⊗&1&5\\\hline1&1&5\\
5&5&1
\end{array}\qquad\begin{array}{c|cc}
⊗&1&7\\\hline1&1&7\\
​7&7&1
\end{array}$

Comments

Then how can A be a group?

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Sorry I can't get your question.Actual question is to prove A is not a group.I also proved it.Is there is any mistake?

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Sorry. i misread the question and didn't see the A, B separation in answer.

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it's ok :)

Answer 2

A- see best answer

B-In Set S numbers which are Co-prime to 8 are 3,5,7 so inverse exist of these numbers and 1 is identity while 0,2,4,6 are not co-prime to 8 so inverse does not exist so we can form pair (1,3)(1,5)(1,7) and these pair form  group under multiplication modulo

Answer 3

({0,1},⊗) is not a group  because : (0e)mod8=0 is haing e=0 as identity but (1e)mod8=1 should have 1 as identity so no single identity is there.

({1,7}, ⊗ ) :

   1. closed

   2. associative

   3. Identity = 1

   4. inerse:  1^-1=7 , 7^-1=1

can someone suggest 2 more subgroups.

Comments

{1,3,5,7} with order 4

with order 2 no further subgroup exist

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Inverse of 7 is 7

The other two groups are {1,3}, {1,5}

Answer 4

(a)In S we have identity element e=1 for multiplication modulo 8 operation.

For Inverse(2) we have to find number x from set s such that  (2x)mod8=1(e).

 Now as you can see 2x will be of ther form 8x+1(then only remainder 1 possible).

8x+1 will always be odd number and any nymber multiplied by 2 will always be even. Therefore no such x exists. Inverse(2)=DNE.

Therefore not Group (We could also have used co prime logic in this group but i like this method).

(b)Moreover for 3 groups we need 1 in each group beacuse it is the identity element for the operation.

Therfore the groups are:-

{1,3}   Inverse(3)=3

{1,5} Inverse(5)=5

{1,7} Inverse(7)=7

All  3 Structure follow closure,Aassociative,e=1, and inverse property.Therfore these 3 are groups and subset of s Therefore subgroups